Pointwise Convergence of a Uniformly Bounded Sequence of Functions

I have a sequence of functions $$f_n:[0,1] \rightarrow \mathbb{R}^n$$ such that $$f_n$$ is uniformly bounded, i.e. $$\|f_n\|\leq M$$ with $$M$$ independent of $$n$$.

Is it true that that the sequence $$(f_n)$$ converges pointwise to some function $$f$$, $$\lim f_n(x)=f(x)$$ for all $$x \in [0,1]$$ ?

2 Answers

It is not true. For example, let the range just be $$\mathbb{R}$$, and take $$f_n(x)=(-1)^n.$$ This is a uniformly bounded sequence, but it does not converge pointwise. You can extend this to $$\mathbb{R}^m$$ in the obvious way.

• True, does this also work if we extract a subsequence of functions? – Lucas27 Aug 24 '19 at 18:39
• I believe that $f_n(x)=\sin nx$ will give you a negative answer. – cmk Aug 24 '19 at 18:44
• @cmk I belive as well that the example with the sin is a counterexample, but I have a hard time proving it. Do you have an idea? – Severin Schraven Aug 24 '19 at 19:18
• – cmk Aug 25 '19 at 17:53
• Thanks! The first one is such a beautiful proof. – Severin Schraven Aug 25 '19 at 19:49

If you want that no subsequence converges, you might define $$f_n(x)=a_n$$ where $$x=a_0.a_1 a_2 a_3\dots$$ is the decimal expansion (with the usual restrictions that renders the expansion unique). Pick your favourite subsequence $$(f_{n_k})_{k\geq 1}$$. Then we consider $$x=a_0.a_1\dots$$ such that $$a_{n_k}=1+(-1)^k$$ and $$a_j=0$$ otherwise. then we have $$f_{n_k}(x)=1+(-1)^k$$ which does not converge.